A Level Mathematics 9709/13 May/June 2025 Past Paper with Answers

Question 1

Find the equation of the tangent to the curve $y = 4x + \dfrac{32}{x^2}$ at the point $(-2, -16)$. What is the $y$-intercept of this tangent?

1. $8$
2. $-8$
3. $5$
4. $-7$

Question 2

The first two terms of a geometric progression are $3\sin^2\alpha$ and $6\sin^3\alpha$, where $0 < \alpha < \dfrac{\pi}{6}$. Given that $S_\infty = \dfrac{1}{4}$, find $\sin\alpha$.

1. $\dfrac{1}{6}$
2. $\dfrac{1}{4}$
3. $\dfrac{1}{3}$
4. $\dfrac{1}{2}$

Question 3

Given that $\displaystyle\int_1^3 \left(\dfrac{b}{(2x-1)^2} + 4\right)\,\mathrm{d}x = 14$, find the value of the constant $b$.

1. $15$
2. $10$
3. $36$
4. $20$

Question 4 · Answer A

Step 1: The general term is $\binom{5}{r}(2)^{5-r}\left(-\dfrac{3x}{2}\right)^r$. Step 2: For $x^2$, set $r = 2$: $\binom{5}{2}(2)^3\left(-\dfrac{3}{2}\right)^2 = 10 \times 8 \times \dfrac{9}{4} = 10 \times 18 = 180$. Step 3: The coefficient of $x^2$ is $180$ (positive, since $(-3/2)^2$ is positive).

Question 5 · Answer A

Step 1: We need $2 - \dfrac{3x}{2} = 1.985$. Step 2: $\dfrac{3x}{2} = 2 - 1.985 = 0.015$. Step 3: $x = \dfrac{0.015 \times 2}{3} = \dfrac{0.03}{3} = 0.01$.

Question 6 · Answer A

Step 1: Replace $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$: $\dfrac{5\sin^2\theta}{\cos\theta} = 2 + 7\cos\theta$. Step 2: Multiply by $\cos\theta$: $5\sin^2\theta = 2\cos\theta + 7\cos^2\theta$. Step 3: Use $\sin^2\theta = 1 - \cos^2\theta$: $5 - 5\cos^2\theta = 2\cos\theta + 7\cos^2\theta$. Step 4: Rearrange: $12\cos^2\theta + 2\cos\theta - 5 = 0$. Step 5: Discriminant $= 4 + 240 = 244 > 0$, giving two real values of $\cos\theta$: $\cos\theta = \dfrac{-2 \pm \sqrt{244}}{24}$. Step 6: $\cos\theta \approx 0.568$ or $\cos\theta \approx -0.734$. Both satisfy $|\cos\theta| \le 1$. Step 7: Each value gives two solutions in $-180^\circ < \theta < 180^\circ$ (since $\cos\theta = k$ gives $\theta = \pm\cos^{-1}(k)$). Step 8: Total solutions $= 4$.

Question 7 · Answer A

Step 1: $N$th term: $a + 2(N-1) = 55$, so $a = 57 - 2N$. Step 2: Sum of first $3N$ terms: $\dfrac{3N}{2}(2a + 2(3N-1)) = 5760$, so $3N(a + 3N - 1) = 5760$. Step 3: Substitute $a = 57 - 2N$: $3N(57 - 2N + 3N - 1) = 3N(56 + N) = 5760$. Step 4: $N^2 + 56N = 1920$, so $N^2 + 56N - 1920 = 0$. Step 5: $(N + 80)(N - 24) = 0$, giving $N = 24$ (rejecting $N = -80$). Step 6: $a = 57 - 2(24) = 57 - 48 = 9$.

Question 8 · Answer A

Step 1: $y$ is decreasing when $\dfrac{\mathrm{d}y}{\mathrm{d}x} < 0$, i.e., $2x^2 + 7x - 15 < 0$. Step 2: Factorise: $(2x - 3)(x + 5) < 0$, giving critical values $x = \dfrac{3}{2}$ and $x = -5$. Step 3: Since the coefficient of $x^2$ is positive, the parabola opens upward, so the expression is negative between the roots. Step 4: Therefore $-5 < x < \dfrac{3}{2}$.

Question 9 · Answer A

Step 1: Factorise the derivative: $3x^2 + 10x - 8 = (3x - 2)(x + 4) = 0$, so $x = \frac{2}{3}$ or $x = -4$. Step 2: The maximum is at $x = -4$ (check second derivative or signs). Step 3: At $x = -4$: $y = (-4)^3 + 5(-4)^2 - 8(-4) + c = -64 + 80 + 32 + c = 48 + c$. Step 4: $48 + c = 34$, so $c = -14$.

Question 10 · Answer A

Step 1: Place $A$ at the origin. $B$ is at $(15, 0)$ and $E$ is on $BC$ with $BE = 5$, so $E = (15, 5)$. Step 2: $AE = \sqrt{15^2 + 5^2} = \sqrt{225 + 25} = \sqrt{250}$. Step 3: By symmetry, $AF = \sqrt{250}$ as well, confirming the arc has radius $\sqrt{250}$.

Question 11 · Answer A

Step 1: Arc length $= r\theta = \sqrt{160} \times 0.9273 = 12.649 \times 0.9273 \approx 11.73$. Step 2: $EC = BC - BE = 12 - 4 = 8$ and $FC = DC - DF = 12 - 4 = 8$. Step 3: Perimeter $= 11.73 + 8 + 8 = 27.73 \approx 27.7$.

Question 12 · Answer A

Step 1: Area of square $ABCD = 12^2 = 144$. Step 2: Area of sector $AEF = \dfrac{1}{2}r^2\theta = \dfrac{1}{2}(160)(0.9273) = 74.18$. Step 3: Area of triangle $ABE = \dfrac{1}{2}(12)(4) = 24$. Step 4: Area of triangle $ADF = \dfrac{1}{2}(12)(4) = 24$. Step 5: Shaded area $= 144 - 24 - 24 - 74.18 = 21.82 \approx 21.8$.

Question 13 · Answer A

Step 1: Gradient of $PR = \dfrac{m - 4}{1 - (-11)} = \dfrac{m - 4}{12}$. Step 2: Gradient of $QR = \dfrac{m - (-1)}{1 - 4} = \dfrac{m + 1}{-3}$. Step 3: For perpendicular lines, the product of gradients $= -1$: $\dfrac{(m-4)(m+1)}{-36} = -1$. Step 4: $(m-4)(m+1) = 36$. $m^2 - 3m - 4 = 36$. $m^2 - 3m - 40 = 0$. Step 5: $(m - 8)(m + 5) = 0$, so $m = 8$ or $m = -5$. Step 6: Since $m > 0$, the positive value is $m = 8$.

Question 14 · Answer A

Step 1: The circle has equation $x^2 + y^2 + 2gx + 2fy + c_0 = 0$. Substituting $P(-13, 5)$: $169 + 25 - 26g + 10f + c_0 = 0$, i.e., $-26g + 10f + c_0 = -194$. Step 2: Substituting $Q(5, 1)$: $25 + 1 + 10g + 2f + c_0 = 0$, i.e., $10g + 2f + c_0 = -26$. Step 3: Substituting $R(2, 10)$: $4 + 100 + 4g + 20f + c_0 = 0$, i.e., $4g + 20f + c_0 = -104$. Step 4: Solving the system: centre is $(-4, 3)$, giving the circle $x^2 + y^2 + 8x - 6y - 60 = 0$. Step 5: Gradient of radius from $(-4, 3)$ to $R(2, 10)$: $\dfrac{10-3}{2+4} = \dfrac{7}{6}$. Step 6: Tangent is perpendicular to the radius, so tangent gradient $= -\dfrac{6}{7}$. Step 7: Tangent: $y - 10 = -\dfrac{6}{7}(x - 2)$, giving $6x + 7y - 82 = 0$. Step 8: $a + b + c = 6 + 7 + (-82) = -69$.

Question 15 · Answer A

Step 1: Differentiate: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{48}{(3x-7)^2} + 3$. Step 2: Set $= 0$: $\dfrac{48}{(3x-7)^2} = 3$, so $(3x-7)^2 = 16$. Step 3: $3x - 7 = \pm 4$, giving $x = \dfrac{11}{3}$ or $x = 1$. Step 4: Both values are valid (neither equals $7/3$), so there are $2$ stationary points.

Question 16 · Answer A

Step 1: $\dfrac{\mathrm{d}y}{\mathrm{d}x} = -18(2x-5)^{-2} + 2$. Step 2: $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 72(2x-5)^{-3} = \dfrac{72}{(2x-5)^3}$. Step 3: At $x = 1$: $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{72}{(2-5)^3} = \dfrac{72}{-27} = -\dfrac{8}{3} < 0$. Step 4: Since $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0$, the stationary point at $x = 1$ is a maximum.

Question 17 · Answer A

Step 1: The minimum of $C$ at $(4, 6)$ becomes the maximum of $C_1$ after reflection in the $x$-axis (reflection flips max/min). Step 2: Apply translation to $(4, 6)$: $(4-3, 6+7) = (1, 13)$. Step 3: Apply reflection in the $x$-axis: $(1, -13)$. Step 4: The maximum point of $C_1$ is $(1, -13)$.

Question 18 · Answer A

Step 1: Translation by $\begin{pmatrix} -3 \\ 7 \end{pmatrix}$: replace $x$ with $x + 3$ and add $7$ to the function. Step 2: After translation: $y = \dfrac{9}{2(x+3)-5} + 2(x+3) - 5 + 7 = \dfrac{9}{2x+1} + 2x + 6 - 5 + 7 = \dfrac{9}{2x+1} + 2x + 8$. Step 3: Reflect in $x$-axis: $y = -\left(\dfrac{9}{2x+1} + 2x + 8\right) = -\dfrac{9}{2x+1} - 2x - 8$. Step 4: So $p = -9$, $q = -2$, $r = -8$. $p + q + r = -9 - 2 - 8 = -19$.

Question 19 · Answer A

Step 1: Complete the square: $f(x) = (x + 2a)^2 - 4a^2 + a$. Step 2: The minimum value is $-4a^2 + a = -33$. Step 3: $4a^2 - a - 33 = 0$. Step 4: Factorise: $(4a + 11)(a - 3) = 0$, so $a = 3$ or $a = -\dfrac{11}{4}$. Step 5: The positive value is $a = 3$.

Question 20 · Answer A

Step 1: From $g^{-1}(x) = \sqrt[3]{2x - 4}$, find $g(x)$. Let $y = g^{-1}(x) = (2x-4)^{1/3}$. Then $y^3 = 2x - 4$, so $x = \dfrac{y^3 + 4}{2}$. Thus $g(x) = \dfrac{x^3 + 4}{2}$. Step 2: $g(0) = \dfrac{0 + 4}{2} = 2$. Step 3: $g(g(0)) = g(2) = \dfrac{8 + 4}{2} = 6$. Step 4: $f(g(g(0))) = f(6) = 36 + 24a + a = 36 + 25a$. Step 5: Set equal to $96$: $36 + 25a = 96$, so $25a = 60$, $a = \dfrac{12}{5}$.